Monday, July 25, 2016

SERIES-PARALLEL CIRCUIT


The goal of series-parallel resistor circuit analysis is to be able to determine all voltage drops, currents, and power dissipation in a circuit. The general strategy to accomplish this goal is as follows:
  • Step 1: Assess which resistors in a circuit are connected together in simple series or simple parallel.
  • Step 2: Re-draw the circuit, replacing each of those series or parallel resistor combinations identified in step 1 with a single, equivalent-value resistor. If using a table to manage variables, make a new table column for each resistance equivalent.
  • Step 3: Repeat steps 1 and 2 until the entire circuit is reduced to one equivalent resistor.
  • Step 4: Calculate total current from total voltage and total resistance (I=E/R).
  • Step 5: Taking total voltage and total current values, go back to last step in the circuit reduction process and insert those values where applicable.
  • Step 6: From known resistances and total voltage / total current values from step 5, use Ohm’s Law to calculate unknown values (voltage or current) (E=IR or I=E/R).
  • Step 7: Repeat steps 5 and 6 until all values for voltage and current are known in the original circuit configuration. Essentially, you will proceed step-by-step from the simplified version of the circuit back into its original, complex form, plugging in values of voltage and current where appropriate until all values of voltage and current are known.
  • Step 8: Calculate power dissipation from known voltage, current, and/or resistance values.
This may sound like an intimidating process, but its much easier understood through example than through description.





In the example circuit above, R1 and R2 are connected in a simple parallel arrangement, as are R3 and R4. Having been identified, these sections need to be converted into equivalent single resistors, and the circuit re-drawn:

The double slash (//) symbols represent “parallel” to show that the equivalent resistor values were calculated using the 1/(1/R) formula. The 71.429 Ω resistor at the top of the circuit is the equivalent of R1and R2 in parallel with each other. The 127.27 Ω resistor at the bottom is the equivalent of R3 and R4 in parallel with each other.
Our table can be expanded to include these resistor equivalents in their own columns:
It should be apparent now that the circuit has been reduced to a simple series configuration with only two (equivalent) resistances. The final step in reduction is to add these two resistances to come up with a total circuit resistance. When we add those two equivalent resistances, we get a resistance of 198.70 Ω. Now, we can re-draw the circuit as a single equivalent resistance and add the total resistance figure to the rightmost column of our table. Note that the “Total” column has been relabeled (R1//R2—R3//R4) to indicate how it relates electrically to the other columns of figures. The “—” symbol is used here to represent “series,” just as the “//” symbol is used to represent “parallel.”


Back to our equivalent circuit drawing, our total current value of 120.78 milliamps is shown as the only current here:
Now we start to work backwards in our progression of circuit re-drawings to the original configuration. The next step is to go to the circuit where R1//R2 and R3//R4 are in series:
Since R1//R2 and R3//R4 are in series with each other, the current through those two sets of equivalent resistances must be the same. Furthermore, the current through them must be the same as the total current, so we can fill in our table with the appropriate current values, simply copying the current figure from the Total column to the R1//R2 and R3//R4 columns:
Now, knowing the current through the equivalent resistors R1//R2 and R3//R4, we can apply Ohm’s Law (E=IR) to the two right vertical columns to find voltage drops across them:

Because we know R1//R2 and R3//R4 are parallel resistor equivalents, and we know that voltage drops in parallel circuits are the same, we can transfer the respective voltage drops to the appropriate columns on the table for those individual resistors. In other words, we take another step backwards in our drawing sequence to the original configuration, and complete the table accordingly:

Finally, the original section of the table (columns R1 through R4) is complete with enough values to finish. Applying Ohm’s Law to the remaining vertical columns (I=E/R), we can determine the currents through R1, R2, R3, and R4 individually:

Having found all voltage and current values for this circuit, we can show those values in the schematic diagram as such:


As a final check of our work, we can see if the calculated current values add up as they should to the total. Since R1 and R2 are in parallel, their combined currents should add up to the total of 120.78 mA. Likewise, since R3 and R4 are in parallel, their combined currents should also add up to the total of 120.78 mA. You can check for yourself to verify that these figures do add up as expected.
A computer simulation can also be used to verify the accuracy of these figures. The following SPICE analysis will show all resistor voltages and currents (note the current-sensing vi1, vi2, . . . “dummy” voltage sources in series with each resistor in the netlist, necessary for the SPICE computer program to track current through each path). These voltage sources will be set to have values of zero volts each so they will not affect the circuit in any way.

15 comments:

  1. I can't understand the statement in how did you get the answers, can you show me what is the formula in getting the resistance the voltage and the current? thanks for the response! :)

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    1. In getting resistance of the given example, you can use the formula in parallel circuit which is 1/RT = 1/R1 + 1/R2 or the other way to make it more understandable R12 = R1xR2 / R1 + R2. In getting the voltage, since the total voltage is given which is 24 volts, to get the voltage drop in V1, V2, V3, V4 simply use the formula of V = I x R. And in getting the current use the formula I = V / R. Thank you for asking! Godbless.

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  2. what is better to use is it parrallel or series? And why?

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    Replies
    1. It depends what you want to do in the context of the circuit. But if in house wiring it is better to used parallel circuit. Because if one appliance is turned on or off it should not affect the other appliances. If houses were wired in series you would have to run everything at the same time, cooker, heater, iron, television etc but with parallel circuits you can select what to turn off and on using switches.

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  3. Hi Ma'am Mylene! Does Kirchoff's law applied in dealing with series-parallel circuits?

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    Replies
    1. No it cannot be applicable in series-parallel circuits but you can use it in series circuit and parallel circuit. Kirchhoff's Current Law can be applied in parallel circuit in getting the Current IT = I1 + I2 + I3 while the Kirchhoff's Voltage Law can be applied in series circuit in getting the Voltage VT = V1 + V2 + V3.

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  4. When do I use the series-parallel circuit?

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    Replies
    1. You may use series-parallel circuit in complicated circuits . For example, combining different logic gates in the circuit like decoder and multiplexer. And circuits having many inputs and outputs.

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  5. lucinda@mail.postmanllc.net

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